\newproblem{lay:1_1_15}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.1.15}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Determine whether the follosing system is consistent (do not fully solve the system).
	\begin{center}
		$\begin{array}{rrrrcr}
			x_1 & -6x_2 &       &       & = &  5 \\
			    &   x_2 & -4x_3 & + x_4 & = &  0 \\
		 -x_1 & +6x_2 & + x_3 & +5x_4 & = &  3 \\
		      &  -x_2 & +5x_3 & +4x_4 & = &  0
		\end{array}$
	\end{center}
}{
   % Solution
	Let us construct the augmented system matrix
	\begin{center}
		$\left(\begin{array}{rrrr|r}
		   1 & -6 &  0 &  0 &  5 \\
		   0 &  1 & -4 &  1 &  0 \\
		  -1 &  6 &  1 &  5 &  3 \\
		   0 & -1 &  5 &  4 &  0 \\
		\end{array}\right)$
	\end{center}
	Now, we apply row operations to solve it
	\begin{center}
		\begin{tabular}{cc}
			 $\begin{array}{c}\mathbf{r}_3\leftarrow \mathbf{r}_3+\mathbf{r}_1\end{array}$ &
				$\left(\begin{array}{rrrr|r}
					 1 & -6 &  0 &  0 &  5 \\
					 0 &  1 & -4 &  1 &  0 \\
					 0 &  0 &  1 &  5 &  8 \\
					 0 & -1 &  5 &  4 &  0 \\
				\end{array}\right)$ \\
			 $\mathbf{r}_4\leftarrow \mathbf{r}_4+\mathbf{r}_2$ &
				$\left(\begin{array}{rrrr|r}
					 1 &  0 & 3 &  0 &  2 \\
					 0 &  1 & 0 & -3 &  3 \\
					 0 &  0 & 1 &  5 &  8 \\
					 0 &  0 & 1 &  5 &  0 \\
				\end{array}\right)$ \\
			 $\mathbf{r}_4\leftrightarrow \mathbf{r}_4-\mathbf{r}_3$ &
				$\left(\begin{array}{rrrr|r}
					 1 &  0 & 3 &  0 &  2 \\
					 0 &  1 & 0 & -3 &  3 \\
					 0 &  0 & 1 &  5 &  8 \\
					 0 &  0 & 0 &  0 & -8 \\
				\end{array}\right)$ \\
		\end{tabular}
	\end{center}
  The system is incompatible since the last row implies the equation $0=-8$.
}
\useproblem{lay:1_1_15}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
